Solutions Manual for Fundamentals of Chemical Engineering Thermodynamics Themis Matsoukas Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Note to the Instructor An effort was made to update all solutions requiring steam tables to conform with the tables in Appendix E of the book, which are based on IAPWS95) It is possible, however, that some problems may make use of older tables Be alert as such discrepancies could confuse students even though the final answers are not much different Please report any mistakes or typos to Themis Matsoukas: matsoukas@psu.edu The author and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein Visit us on the Web: InformIT.com/ph Copyright © 2013 Pearson Education, Inc This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials ISBN-10: 0-13-269320-8 ISBN-13: 978-0-13-269320-2 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Contents Introduction Phase Diagrams of Pure Fluids 13 Energy & the First Law 59 Entropy & the Second Law 103 Calculation of Properties 125 Balances in Open Systems 143 VLE of pure Fluid 193 Phase Behavior of Mixtures 223 Properties of Mixtures 245 10 VLE of Mixture 271 11 Ideal Solution 291 12 Non-Ideal Solution 311 13 Miscibility, Solubility and other Phase Equilibria 347 14 Reactions 371 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Introduction This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute I NTRODUCTION Problem 1.1 Solution We will need the following unit conversions: ft D 0:3048 m; lb D 0:454 kg; lbmol D 454 mol (Note on the conversion mol to lbmol: one mol has a mass equal to the molecular weight in g, while one lbmol has a mass equal equal to the molecular weight in lb.) We also need the molar mass of ammonia which is Mm D 17 g/mol D 17 10 kg/mol a) Specific Volume: V D D 0:02421 ft3 /lb D 0:00151 m3 /kg D 1:51 cm3 /g 41:3 lb/ft3 b) Molar Volume Vmolar D VMm D 2:567 10 m3 /mol D 25:67 cm3 /mol D 0:4116 ft3 /lb-mol This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 1.2 Solution First we write the given equation as TK D A B ln Pbar /=.ln 10/ C where TK refers to T in kelvin, Pbar refers to pressure in bar, and the logarithm is natural Next we use TF D 1:8.TK 273:15/ C 32 and Pbar D Ppsi 14:5 and substitute these values into the above equation After some manipulation the result is TF D B.1:8/.ln 10/ A ln 10 C ln.14:5/ ln Ppsi C.1:8/ 1:8/.273:15/ C 32 Doing the algebra, TF D 5062:37 13:2153 ln Ppsi 302:217 Therefore, A0 D 13:2153; B D 5062:37 C D 302:217 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute I NTRODUCTION Problem 1.3 Solution a) The mean velocity is  vN D where kB D 1:38 10 23 8kB T m Ã1=2 J=K, T D 273:16 K The mass of the water molecule is 18 10 kg=mol Mm D D 2:98904 NAV 6:022 1023 mol mD 10 26 kg: The mean velocity is vN D 566:699 m=s D 2040:12 km=h D 1267 mph This result depends only on temperature and since all three phases are the same temperature, molecules have the same mean velocity in all three phases b) The mean kinetic energy is EN kin D 12 mv D 21 mv where v is the mean squared velocity, v2 D 3kB T : m With this the mean kinetic energy is EN kin D kB T D 5:65441 10 21 J This is the mean kinetic energy per molecule in all phases The number of molecules in kg of water is N D kg 6=022 10 kg=mol 18 1023 D 3:34556 1025 The total kinetic energy in kg of water at 0.01 ı C (regardless of phase) is EN kin D 5:65441 10 21 J 3:34556 1025 D 189; 171 J D 189 kJ Comment: This is the translational kinetic energy of the molecule, i.e., the kinetic energy due to the motion of the center of mass A water molecule possesses additional forms of kinetic energy that arise from the rotation of the molecule, the bending of bonds, and the vibration of oxygen and hydrogen atoms about their equilibrium positions These are not included in this calculation as the Maxwell-Boltzmann distribution refers specifically to the translational kinetic energy c) The above calculation shows that the mean kinetic energy depends only on temperature (it is independent of pressure or of the mass of the molecule) Therefore, oxygen at 0.01 ı C has the same kinetic energy as water at the same temperature: EN kin D kB T D 5:65441 10 21 J This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 1.4 Solution Potential 22 10 22 Joule 10 10 22 10 22 10 r ˚ or so To determine this value accurately we must set the a) The potential has a minimum just above A derivative of the potential equal to zero solve for the value of r To this easily, we define a new variable x D r= and rewrite the potential as: ˆ D a x 12 x By chain rule we now have: dF dF dx D D dr dx dr 12x 13 C 6x dx dr Setting this to zero and solving for x we have: 12x 13 C 6x D0 ) x D 21=6 Since r D x, the value of r that minimizes the potential is ˚ D 4:25 A ˚ r D 21=6 D 4:24964/.3:786 A/ b) If we imagine N molecules to be situated at the center of cubes whose sides are equal to r , L L the volume occupied is V D N.r /3 These N molecules correspond to N=NAv mol and their total mass is M D N Mm NAV where Mm is the molar mass of methane (Mm D 16 10 kg/mol) For the density, therefore, we obtain the following final formula: Mm D NAV r /3 By numerical substitution we finally obtain the density: D 16 10 kg/mol ˚ 6:024 1023 mol /.4:25 A/.10 10 ˚ m/A/ D 346 kg/m3 D 0:346 g/cm3 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute I NTRODUCTION c) Specific volumes of saturated liquid methane are listed in Perry’s Handbook from which we can compute the densities We notice (as we would have expected) that liquid volumes near the critical point (Tc D 190:55 K) vary with pressure, from 162.3 kg/m3 at the critical point to 454 kg/m3 around 90 K Our value corresponds to Perry’s tabulation at about 160 K Our calculation is approximate and does not incorporate the effect of pressure and temperature Notice that if we pick a distance somewhat different from r the result will change quite a bit because if the third power to which this distance is raised But the important conclusion is that the calculation placed the density right in the correct range between the lowest and highest values listed in the tables This says that our molecular picture of the liquid, however idealized, is fairly close to reality 10 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P HASE D IAGRAMS OF P URE F LUIDS Problem 2.22 Solution a) Vapor b) At 0.1 bar, 200 C, o-xylene is essentially in the ideal-gas state (why?) V D RT D 0:393377 m3 =mol P The volume of the tank is V tank D 200 mol/.0:393377 m3 =mol/ D 39:3 m3 : c) At 44.9 bar, 200 C, the reduced temperature an pressure is Tr D 0:75; Pr D 1:2 From the Lee-Kesler tables we find Z D 0:84435; The molar volume is V2 D Z D 0:0453; Z D 0:8585 ZRT D 0:000752133 m3 =mol P2 and the number of moles nD V tank D 52301:5 mol V2 44 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute (2.2) Problem 2.23 Solution a) We first calculate the coefficient of thermal expansion form the empirical equation given above:  à  à @V @t ˇD V @t P @T where t stands for temperature in celsius and T for temperature in kelvin Using the polynomial expression given in the problem statement we find a1 C 2a2 t C 3a3 t C a1 t C a2 t C a3 t ˇD For constant pressure process, dV D ˇd T V ) V2 D e ˇ T2 V1 T1 / For this calculation we will use an average value of ˇ between 18 ı C and 40 ı C: ˇav D 0:5.ˇ18 C ˇ40 / D 0:5.7:09 10 C 7:27 10 /K D 7:18 10 K With this value, the change in volume is V2 D 1:016 V1 or an increase of 1/6% Note: We have treated ˇ as nearly constant To determine the validity of this assumption we make a graph of ˇ in the temperature range of interest: 0.0010 Β 1K 0.0008 0.0006 0.0004 0.0002 0.0000 20 25 30 35 40 T °C The coefficient ˇ changes very slowly with temperature, therefore the use of an average value is justified b) We start with dV D ˇd T ÄdP: V Assuming the contained to be rigid, volume remains constant, i.e., dV D We then integrate the above equation and solve for P : D ˇd T ÄdP ) Numerical substitutions: P D 7:18 ÄP D ˇT ) P D ˇT Ä 10 K /.22 K/ D 287 bar 52 10 bar 45 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P HASE D IAGRAMS OF P URE F LUIDS Problem 2.24 Solution The numerical results are summarized in the atatched tables: 46 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P=70 bar Fluid Tc Pc w CO2 304.129 73.74 0.225 K bar T P Psat Phase 293.15 70 57.2 Liquid K bar bar a 0.381864 b 0.0000297087 A B 0.449994 0.085326 Z roots 0.186938 Z 0.186938 V 0.0000650879 Meter3 Mol moles mass 7681.92 338.081 Mol kg T P Psat Phase 293.15 60 57.2 Liquid K bar bar a 0.381864 b 0.0000297087 A B 0.385709 0.0731366 Z roots 0.169418 0.338098 0.492484 Z 0.169418 V 0.000068819 Meter3 Mol moles mass 7265.44 319.752 Mol kg Joule Meter3 Mol2 Meter3 Mol P=60 bar Joule Meter3 Mol2 Meter3 Mol This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute solution-2-Table.nb 50 bar T P Psat Phase 293.15 50 57.2 Vapor K bar bar a 0.381864 b 0.0000297087 A B 0.321424 0.0609471 Z roots 0.160909 0.18658 0.652512 Z 0.652512 V 0.000318067 Meter3 Mol moles mass 1572 69.1836 Mol kg Joule Meter3 Mol2 Meter3 Mol This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 2.25 Solution First we collect the parameters for isobutane: Tc D 408:1 K; Pc D 36:48 bar; ! D 0:181; Mw D 58:123 10 kg/mol We are given T D 294:26 K, P D 4:13793 bar With this information we find that the compressibility equation has three real roots: Z1 D 0:0185567; Z2 D 0:100577; Z3 D 0:880866; We know that the phase is liquid (since the given temperature is below the saturation temperature at the given pressure), therefore the correct compressibility factor is the smallest of the three: Z D 0:0185567 The corresponding molar volume is V DZ 8:314 J/mol K/.294:26 K/ RT D 0:0185567/ D 1:09713 P 4:13793 105 Pa 10 m3 /mol The amount (moles) of isobutane is nD 5000 kg D 86024:5 mol 58:123 10 kg/mol Therefore, the volume of the tank is V tank D 86024:5 mol/.1:09713 10 m3 /mol/ D 9:44 m3 a) Before we solve the problem it is useful to look at the P V graph first P2 P1 V Since the volume of the tank and the mass of isobutane remain the same, the molar volume also stays the same In other words, the new state must be on the vertical line that passes through the initial state Since temperature is higher, the final state will be above state (marked as state in the above figure) This is somewhat surprising: one might think that some vapor may be generated since temperature increases Instead, the system moves firther into the compressed liquid region! This is because heating takes place under constant volume.1 If heating were to take place under constant pressure insead, the final state would move to the right of state 1, possibly creating some vapor 49 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P HASE D IAGRAMS OF P URE F LUIDS b) Since the total volume and mass in the tank remain the same, the molar volume must also stay the same, namely, V D 1:09713 10 m3 /mol The pressure can now be calculated directly from the SRK equation: P D RT V b a V V C b/ Notice, however, that the parameter a must be recalculated because it depends on temperature With T D 308:15 K we find a D 1:68988 J m3 /mol2 Using this value of a, the previous value of b, and V D 1:09713 10 m3 /mol, the SRK equation gives P D 70:1 bar This represents an increase of 66 bar even though temperature increased only by 20 ı F! The reason is that isotherms in the compressed liquid state are very steep, resulting in large pressure change under constantvolume heating 50 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 2.26 Solution a) At 30 ı C, bar the SRK equation has the following three real roots Since the phase is vapor (why?) we pick the largest root: ZRT Z D 0:977286 V D D 0:0246314 m3 =mol P b) At 30 ı C, 10 bar the SRK equation has the following three real roots The phase is liquid (why?), therefore we pick the smallest root: Z D 0:0413973 V D ZRT D 0:000104337 m3 =mol P b) At 30 ı C, 4.05 bar the SRK equation has the following three real roots Since the system is saturated, the smallest root is the liquid and largest is the vapor: ZL D 0:0413973; VL D ZL RT D 0:000104698 m3 =mol P sat ZV D 0:901582; VV D ZV RT D 0:00561071 m3 =mol P sat The literature values from the NIST Web Book are VL D 0:00010678 m3 =mol VV D 0:0055461 m3 =mol The SRK values are off by 2% (liquid) and 1:2% (vapor) These errors are pretty small 51 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P HASE D IAGRAMS OF P URE F LUIDS Problem 2.27 Solution 50 150 °C 40 T Tc c P bar 30 20 sat Liq sat Vap 10 30 °C 10 10 0.001 0.005 0.010 0.050 V m3 mol Note: For maximum readability use semilog coordinates (linear axis for pressure, logarithmic axis for volume) 52 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 2.28 Solution The the general form for the differential of V is dV D ˇd T V ÄdP Using the given expressions for ˇ and Ä we have d ln V D dT T dP P ŒA Integration is this differential from V0 , T0 , P0 to V , T , P is very simple in this case because the variables happen to be separated (each of the three terms contains one variable only) The result is ln V T D ln V0 T0 ln P P0 ŒB The same result is obtained if we adopt an arbitrary integration path, say from T0 , P0 , under constant T to T0 , P , and then under constant P to T , P As we can easily verify, the differential of the above is indeed equal to Eq [A] Equation [B] can be rearranged to write  à V T P0 D ln ln V0 T0 P or P0 V0 PV D T T0 In other words we have obtained the ideal-gas law Based on the final result we can certainly say that this equation of state is not appropriate for liquids Even before integration, however, we could reach the same conclusion by looking at the T and P dependence of ˇ and Ä The inverse dependence of ˇ on T (and of Ä on P ) indicates that these parameters vary quite a bit with pressure and temperature This is a characteristic of gases The values of Ä and ˇ for solids and liquids are typically small numbers and vary much less with temperature and pressure 53 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P HASE D IAGRAMS OF P URE F LUIDS Problem 2.29 Solution Using the hint we write:  à @V ÄD D V @P T  V Ã. @P @V à T Starting with the SRK equation, P D the derivative wrt V is:  @P @V à D T RT V b a V V C b/ a b C 2V V b C V /2 RT b V /2 Numerical substitutions T D 303:15 P D 4:05 a D 1:60525 b D 0:000075256 K bar J=m3 =mol2 m3 =mol ZL D 0:0168238 VL D 0:000104698 @P =@V /T D 1:62039 Ä L D 0:000589445 1012 m3 =mol bar mol=m3 bar ZV D 0:901582 VV D 0:00561071 @P =@V /T D 6:44372 Ä V D 0:276596 107 m3 =mol bar mol=m3 bar 54 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 2.30 Solution a) By definition, ˇD V  @V @T à P The partial derivative will be approximated as a finite difference between two states A and B at the same pressure: à  VA VB @V @T P TA TB For V we must use a value between VA and VB Choosing V D VA C VB /=2, the final result is VA VA C VB TA ˇ VB TB Pressure A B const P D C Volume a) At 25ı C, bar, the system is compressed liquid Assuming the liquid to be incompressible, the required volumes are those of the saturated liquid TA TB D 20 ı C VA D 30 ı C VB D 0:001002 m3 =kg D 0:001004 m3 =kg The coefficient of isothermal compressibility is ˇ D 1:99402 10 K b) The answer at 10 bar is the same because the assumption of incompressibility implies that the isotherms are vertical and the molar volumes the same as in the previous part c) In this case the state is superheated vapor We select two temperatures around 200 ı C and apply the same procedure: TA D 150 ı C VA D 1:9367 m3 =kg TB D 250 ı C VB D 2:4062 m3 =kg and we find ˇ D 2:16215 10 K 55 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute P HASE D IAGRAMS OF P URE F LUIDS Problem 2.31 Solution Solution Assuming isotherms in the compressed liquid region to be vertical, ˇ is calculated as  à @V V2 V1 ˇD D V CV T V @T P T1 2 where T1 , T2 are two temperatures around 24 ı C, and V1 , V2 , are the volumes of the saturated liquid at these temperatures, to be calculated using the Rackett equation With T1 D 20ı C, T2 D 30 ı C we find ˇ D 1:475 10 K Calculation are shown in the attached notebook 56 This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Rackett-Ethanol.nb Rackett Equation Data for ethanol In[6]:= Tc = 513.9 K; Pc = 61.48 bar; w = 0.645; Vc = 167 cm3 ê mol; Zc = 0.24; Rackett Equation In[5]:= V@t_D := Vc ZcH1-têTcL 0.2857 Calculations In[14]:= In[29]:= T1 = H20 + 273.15L K; T2 = H30 + 273.15L K; V1 = V@T1D V2 = V@T2D V2 - V1 b = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ; HV1+V2L ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ T2 - T1 Print@"b = ", bD Out[29]= 54.4331 cm3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ mol Out[30]= 55.2419 cm3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ mol 0.00147493 b = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ K This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute This text is associated with Themis Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute ... kinetic energy of the molecule, i.e., the kinetic energy due to the motion of the center of mass A water molecule possesses additional forms of kinetic energy that arise from the rotation of the molecule,... Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 2.4 Solution Solution Initial... Matsoukas, Fundamentals of Chemical Engineering Thermodynamics 0-13-269306-2 (978-0-13-269306-6) Copyright © 2013 Pearson Education, Inc Do not redistribute Problem 2.10 Solution Solution The